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Contests
National and Regional Contests
South Africa Contests
South Africa National Olympiad
2002 South africa National Olympiad
2002 South africa National Olympiad
Part of
South Africa National Olympiad
Subcontests
(6)
6
1
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WEIRD equation with rationals
Find all rational numbers
a
a
a
,
b
b
b
,
c
c
c
and
d
d
d
such that
8
a
2
−
3
b
2
+
5
c
2
+
16
d
2
−
10
a
b
+
42
c
d
+
18
a
+
22
b
−
2
c
−
54
d
=
42
,
8a^2 - 3b^2 + 5c^2 + 16d^2 - 10ab + 42cd + 18a + 22b - 2c - 54d = 42,
8
a
2
−
3
b
2
+
5
c
2
+
16
d
2
−
10
ab
+
42
c
d
+
18
a
+
22
b
−
2
c
−
54
d
=
42
,
15
a
2
−
3
b
2
+
21
c
2
−
5
d
2
+
4
a
b
+
32
c
d
−
28
a
+
14
b
−
54
c
−
52
d
=
−
22.
15a^2 - 3b^2 + 21c^2 - 5d^2 + 4ab +32cd - 28a + 14b - 54c - 52d = -22.
15
a
2
−
3
b
2
+
21
c
2
−
5
d
2
+
4
ab
+
32
c
d
−
28
a
+
14
b
−
54
c
−
52
d
=
−
22.
5
1
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Inscribed semicircles
In acute-angled triangle
A
B
C
ABC
A
BC
, a semicircle with radius
r
a
r_a
r
a
is constructed with its base on
B
C
BC
BC
and tangent to the other two sides.
r
b
r_b
r
b
and
r
c
r_c
r
c
are defined similarly.
r
r
r
is the radius of the incircle of
A
B
C
ABC
A
BC
. Show that
2
r
=
1
r
a
+
1
r
b
+
1
r
c
.
\frac{2}{r} = \frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c}.
r
2
=
r
a
1
+
r
b
1
+
r
c
1
.
4
1
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1000000 is the product of 3 integers > 1
How many ways are there to express 1000000 as a product of exactly three integers greater than 1? (For the purpose of this problem,
a
b
c
abc
ab
c
is not considered different from
b
a
c
bac
ba
c
, etc.)
3
1
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Funny problem with squares
A small square
P
Q
R
S
PQRS
PQRS
is contained in a big square. Produce
P
Q
PQ
PQ
to
A
A
A
,
Q
R
QR
QR
to
B
B
B
,
R
S
RS
RS
to
C
C
C
and
S
P
SP
SP
to
D
D
D
so that
A
A
A
,
B
B
B
,
C
C
C
and
D
D
D
lie on the four sides of the large square in order, produced if necessary. Prove that
A
C
=
B
D
AC = BD
A
C
=
B
D
and
A
C
⊥
B
D
AC \perp BD
A
C
⊥
B
D
.
2
1
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Geometric progression with sum 111
Find all triples of natural numbers
(
a
,
b
,
c
)
(a,b,c)
(
a
,
b
,
c
)
such that
a
a
a
,
b
b
b
and
c
c
c
are in geometric progression and
a
+
b
+
c
=
111
a + b + c = 111
a
+
b
+
c
=
111
.
1
1
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Easy geometry
Given a quadrilateral
A
B
C
D
ABCD
A
BC
D
such that
A
B
2
+
C
D
2
=
A
D
2
+
B
C
2
AB^2 + CD^2 = AD^2 + BC^2
A
B
2
+
C
D
2
=
A
D
2
+
B
C
2
, prove that
A
C
⊥
B
D
AC \perp BD
A
C
⊥
B
D
.